@JosuéMolina: Glad to help. One last example, notice that by taking the determinant of each side of the first equation we get Cassini's identity, $f_{n+1}f_{n-1} - f_n^2 = (-)^n$.
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user26872Apr 25 '12 at 14:57

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There is actually an interesting article about this interpretation of the Fibonacci sequence in this month's Mathematics Magazine for those who are curious about some more of its utility in proofs.
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Alexander LApr 26 '12 at 3:17

How did you obtain that? Did you know it beforehand?
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JosuéApr 25 '12 at 5:46

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No. I reasoned what $f_{2n}$ had to be if the formula you were asking me to prove were true by using $f_{2n + 1} = f_{2n} + f_{2n-1}$ and writing $2n - 1 = 2(n-1) + 1$.
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user29743Apr 25 '12 at 5:48

Just a remark - when you go to write out the induction, the odd case isn't bad, but the even case is equivalent to the formula $2f_{n+1}^2 + f_n^2 - f_{n-1}^2 = f_{n+2}^2 - f_n^2$ which follows from expanding out $f_{n+2}$ and one (but not both) copy of $f_{n+1}$ using the recurrence.
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user29743Apr 25 '12 at 6:04

Sorry for the comment proliferation. By "one copy" I mean write $2f_{n+1}^2 = f_{n+1}^2 + f_{n+1}^2$ and expand only one of those two summands.
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user29743Apr 25 '12 at 6:12

In an effort to find a more beautiful proof, we would try looking at this geometrically. By looking at the golden rectangle in terms of area, we can deduce that $f_{n+1}^2+f_{n}^2=f_{n+2}f_{n+1}-f_{n}f_{n-1}$. From the golden rectangle we can also get: $f_{n+1}^2-f_{n-1}^2=f_{n+2}f_{n+1}+f_{n}f_{n+1}$. Perhaps something pretty can come of these?

Here's a geometrical approach that I posted as an answer to a question that was closed as a duplicate of this one. I'll repost it here, I hope that's ok.

First create a rectangle with length $F_{2n+1}$ and height $F_1 = 1$. Then cut this up into two rectangles with sizes $F_{2n} \times F_2$ (i.e. $F_{2n} \times 1$) and $F_{2n-1} \times F_1$ (i.e. $F_{2n-1} \times 1$). Put the latter under the former.

Then cut the resulting figure into rectangles with sizes $F_{2n-1} \times F_3$ and $F_{2n-2} \times F_2$. Again, put the latter under the former.

Now cut the resulting figure into rectangles $F_{2n-2} \times F_4$ and $F_{2n-3} \times F_3$ and put the latter under the former.

Eventually, after $n-1$ steps you will get a figure built up from two squares with sizes $F_{n+1}^2$ and $F_n^2$.

See below figure for an example with $n=6$, i.e. $F_{2n+1} = F_{13} = 233$. The first line is the $F_{13} \times F_1$ rectangle. After $n-1=5$ steps you will reach a figure which can be divided into $F_7^2$ and $F_6^2$ as desired.